Question

In 108 g of water, 18 g of non - volatile compound is dissolved. At 100$^0C$ the vapour pressure of the solution is 750 mm Hg. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in $gmo{l^{ - 1}}$ is :
A. 128
B. 182
C. 152
D. 228

Answer 1

Hint: This is a numerical and the formula here that can lead to determination of molar mass of the given compound can be that of vapour pressure one. We know that -
$\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{{W_2}{M_1}}}{{{W_1}{M_2}}}$
Where ${M_1}$ = Molar mass of the water
${M_2}$ = Molar mass of the compound
${W_1}$= Weight of the water
${W_2}$ = Weight of the compound

Complete step by step solution:
In such types of questions, first, let us see what is given to us and what we need to find out.
We are given :
Mass of water = $108 g$
Mass of non volatile compound = $18 g$
Temperature = 100$^0C$
Vapour pressure of the solution = $750 \text (mm Hg)$
Further assume that the compound does not undergo association or dissociation
To find :
Molar mass of the compound in (g /mol )
We have the formula -
$\dfrac{{{p^0} - p}}{{{p^0}}} = \dfrac{{{W_2}{M_1}}}{{{W_1}{M_2}}}$
Where ${M_1}$ = Molar mass of the water
${M_2}$ = Molar mass of the compound
${W_1}$= Weight of the water
${W_2}$ = Weight of the compound
On putting the values in above equation, we get -
$\dfrac{{760 - 750}}{{760}} = \dfrac{{18 \times 18}}{{108 \times {M_2}}}$
Solving the above, we get
${M_2} = \dfrac{{18 \times 18 \times 76}}{{108}}$
${M_2} = {228 \text gmo{l^{-1}}}$

So, the option (D) is the correct answer.

Note: Many people get confused with given mass and molar mass. These both are different things. The molar mass of a compound is the total mass of the molecule which is formed by the sum of masses of all elements multiplied by the number of atoms of each element. It is fixed while the given mass is the amount of that substance taken at that time during the experiment. It can be any amount required and is thus variable.